Two projectiles A and B thrown with speeds in the ratio 1 : `sqrt(2)` acquired the same heights. If A is thrown at an angle of `45^(@)` with the horizontal, the angle of projection of B will be

Options

A. `0^@`
B. `60^@`
C. `30^@`
D. `45^@`

Answer

To solve the problem, we need to determine the angle of projection of projectile B, given that both projectiles A and B reach the same height, and their speeds are in the ratio of 1:√2.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Speed ratio of A to B: \( \frac{u_A}{u_B} = \frac{1}{\sqrt{2}} \)
- Angle of projection for A: \( \theta_A = 45^\circ \)
- Both projectiles reach the same height.

2. **Use the Formula for Maximum Height:**
The maximum height \( H \) reached by a projectile is given by the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
where \( u \) is the initial speed, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

3. **Set Up the Height Equations:**
For projectile A:
\[ H_A = \frac{u_A^2 \sin^2 \theta_A}{2g} \]
For projectile B:
\[ H_B = \frac{u_B^2 \sin^2 \theta_B}{2g} \]
Since both projectiles reach the same height, we can set \( H_A = H_B \):
\[ \frac{u_A^2 \sin^2 \theta_A}{2g} = \frac{u_B^2 \sin^2 \theta_B}{2g} \]
We can cancel \( 2g \) from both sides:
\[ u_A^2 \sin^2 \theta_A = u_B^2 \sin^2 \theta_B \]

4. **Substitute the Speed Ratio:**
Substitute \( u_A = \frac{1}{\sqrt{2}} u_B \) into the equation:
\[ \left(\frac{1}{\sqrt{2}} u_B\right)^2 \sin^2(45^\circ) = u_B^2 \sin^2 \theta_B \]
Simplifying gives:
\[ \frac{1}{2} u_B^2 \cdot 1 = u_B^2 \sin^2 \theta_B \]
(since \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \) and \( \sin^2(45^\circ) = \frac{1}{2} \))

5. **Cancel \( u_B^2 \):**
Assuming \( u_B \neq 0 \), we can divide both sides by \( u_B^2 \):
\[ \frac{1}{2} = \sin^2 \theta_B \]

6. **Solve for \( \sin \theta_B \):**
Taking the square root of both sides, we get:
\[ \sin \theta_B = \frac{1}{\sqrt{2}} \]

7. **Determine the Angle \( \theta_B \):**
The angle whose sine is \( \frac{1}{\sqrt{2}} \) is:
\[ \theta_B = 30^\circ \]

### Final Answer:
The angle of projection of projectile B is \( 30^\circ \).

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